Finding the derivative of powers using First Principles

Question

I am unsure how to differentiate $f(x)=x^{\frac34}$ using first principles definition of the derivative.

I appreciate any help.

Chris

The ... first ... principles? Do you mean the definition of the derivative? — ajotatxe, Sep 17 '19 at 14:31
Not really, more seeking help in how to apply the first principles with this particular question. — Christopher, Sep 17 '19 at 14:35
It seems that there is confusion with what you mean by first principles. What are these first principles aside from the definition of derivative? — Kitter Catter, Sep 17 '19 at 14:36

Andrew Chin · Accepted Answer · 2019-09-17T15:24:54.450

1

Hint:

The first principles definition of the derivative is given by $$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}.$$

In your case, let $f(x)=x^{\frac34}$. Then, the limit becomes \begin{align} f'(x)&=\lim\limits_{h\to0}\frac{(x+h)^{\frac34}-x^{\frac34}}{h} \end{align}

Observe what happens when we conjugate the numerator: $$ \frac{(x+h)^{\frac34}-x^{\frac34}}{h}\color{blue}{\cdot\frac{(x+h)^{\frac34}+x^{\frac34}}{(x+h)^{\frac34}+x^{\frac34}}}=\frac{(x+h)^{\frac32}-x^{\frac32}}{h\big({(x+h)^{\frac34}+x^{\frac34}}\big)} $$

What would happen if you were to conjugate the numerator again? $$ \frac{(x+h)^{\frac32}-x^{\frac32}}{h\big({(x+h)^{\frac34}+x^{\frac34}}\big)}\color{blue}{\cdot\frac{(x+h)^{\frac32}+x^{\frac32}}{(x+h)^{\frac32}+x^{\frac32}}}=\frac{(x+h)^3-x^3}{h\big({(x+h)^{\frac34}+x^{\frac34}}\big)\big((x+h)^{\frac32}+x^{\frac32}\big)} $$ The numerator can now be expanded, and you should be able to take it from here.

edited Sep 17 '19 at 15:24

answered Sep 17 '19 at 14:45

Andrew Chin

7,389

Sorry, I am having a bit of trouble understanding how you get (x+h)^3/4 when conjugating the numerator – Christopher Sep 17 '19 at 15:04
Using the algorithm for conjugating based on the factorization of a difference of squares: $$(a-b)\color{blue}{(a+b)}=a^2-b^2.$$ Specifically for this example, it is also a good idea to consider that $$(a^m-b^n)\color{blue}{(a^m+b^n)}=a^{2m}-b^{2n}.$$ – Andrew Chin Sep 17 '19 at 15:07
Awesome, that makes sense. Thanks for the reply. I will try figure out the result of an additional conjugation :) – Christopher Sep 17 '19 at 15:13
Strangely, I got a 0 for the numerator, when i conjugated it again. Is that what i should be aiming to do, or would there be some error in my calculation? – Christopher Sep 17 '19 at 15:21
Added the conjugation in an edit to my answer. There must have been some error in your calculation, or you did not write the correct conjugate. – Andrew Chin Sep 17 '19 at 15:25

Dr. Sonnhard Graubner · Answer 2 · 2019-09-17T14:51:47.810

0

Compute $$\lim_{h\to 0}\frac{\sqrt[4]{(x+h)^3}-\sqrt[4]{x^3}}{h}$$ if $h\ne 0$

edited Sep 17 '19 at 14:51

answered Sep 17 '19 at 14:41

Dr. Sonnhard Graubner

95,283

1

Couldn't edit your post, but perhaps you meant $\sqrt[4]{\color{blue}{x}^3}$ instead? – Andrew Chin Sep 17 '19 at 14:49
Thank you, just corrected! – Dr. Sonnhard Graubner Sep 17 '19 at 14:52

score 0 · Answer 3 · answered Sep 17 '19 at 15:43

We Know That The Derivative of $f(x)$ is Given by $f(x+h)-f(x)\over h$

Since $f(x) = x^{3\over 4}$

Derivative is Given By : $f(x+h)^{3\over 4} - f(x)^{3\over 4}\over h$

Using Bionomial Theroem , derivative= $1\over h$ [$(x^{3\over 4 }$ + ${3\over 4} x^{-1\over4}h$ +other terms containing higher power of h ) - $x^{3\over4}$ ]

= $1\over h $[${3\over4} x^{-1\over4}h$ + other Terms containing higher power of h ]

=${3\over4}x^{-1\over4}$ + other Terms containing higher power of h

=${3\over4}x^{-1\over4}$

Hence Derivative of $x^{3\over4}$ Is ${3\over4}x^{-1\over4}$

Finding the derivative of powers using First Principles

3 Answers3